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Old 17th September 2010, 01:00 AM   #1
Irakli is offline Irakli  United States
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Default Help - resistance of the mesh

Hi All,

Please help me solve this problem:
What is the resistance between points A and B. All resistanses in the mesh are equal to R. The mesh size(NxN) is large.

I remember solving this problem when I was in school, but can't do it now

I also remember it has a very elegant solution.

If you know the answer, please give me a hint
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Old 17th September 2010, 03:14 AM   #2
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I remember that one. You draw equipotential lines and, since they are equipotential you can connect them without changing anything. Once you have done that the entire thing becomes trivial.

Since the array is symmetrical, you short the same way on both sides.

Actually the one I recall was a cube of resistors where you measure across the diagonal. This one is similar. So fold the whole thing top over to the bottom, hinging on the A-B line. That leaves resistors only below and all are now R/2 except the ones on the hinge line.

Last edited by bob91343; 17th September 2010 at 03:17 AM.
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Old 17th September 2010, 03:22 AM   #3
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I'll guess at 0.6R
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Old 17th September 2010, 03:52 AM   #4
Irakli is offline Irakli  United States
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Quote:
Originally Posted by bob91343 View Post
I remember that one. You draw equipotential lines and, since they are equipotential you can connect them without changing anything. Once you have done that the entire thing becomes trivial.

Since the array is symmetrical, you short the same way on both sides.

Actually the one I recall was a cube of resistors where you measure across the diagonal. This one is similar. So fold the whole thing top over to the bottom, hinging on the A-B line. That leaves resistors only below and all are now R/2 except the ones on the hinge line.

Cube is easy

with this mesh we can fold it across AB line, but this does not help much
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Old 17th September 2010, 04:46 AM   #5
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I wonder if it helps that you would get the same value no matter which resistor you pick, since the array is infinite. So you can substitute for any resistor the value of the total matrix.

I don't know. This isn't really an electronics problem; it's a logic or math problem.
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Old 17th September 2010, 05:31 AM   #6
wwenze is offline wwenze  Singapore
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Build it?

YouTube - EEVblog #25 - The Infinite Resistor Puzzle
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Old 17th September 2010, 04:38 PM   #7
Irakli is offline Irakli  United States
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Quote:
Originally Posted by wwenze View Post
Build it?
Thanks, always nice to have experimental conformation.
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Old 23rd September 2010, 12:05 AM   #8
Irakli is offline Irakli  United States
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Searched the web intensely and what I thought an "elegant" solution seems to be a very spooky one.

Idea is to pump in current in one node (point A) and pump it out from another one (point B). Total current through A->B supposed to be i/4+i/4=i/2, thus Rtot=R/2. I do not think that i/4 current will flow each direction because of symmetry. Symmetry is broken with the presence of the second node. Symmetry case would work if someone wants to measure resistance between arbitrary node and infinitely remote point.

May be I am missing something.
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